Question: How many different positive three-digit integers can be formed using only the digits in the set $\{2, 3, 5, 5, 5, 6, 6\}$ if no digit may be used more times than it appears in the given set of available digits?
If all three digits are different, there are 4 choices for the first digit, 3 for the second, and 2 for the third, giving $(4)(3)(2) = 24$ integers. If two of them are the same, the repeated digit digit is either 5 or 6. There are 2 choices for the repeated digit, then 3 choices for the non-repeating digit, and 3 ways to arrange these digits (for example, if the repeating digit is 5 and the non-repeating digit is 6, we can have 655, 565, and 556). This gives $(2)(3)(3) = 18$ integers. Finally, if all three digits are the same, the number must be 555. So there are $24+18+1 = \boxed{43}$ possible integers.